It should be uncountable, otherwise it's trivially true that is perfect since it' empty!
Lemma: Let be a second countable metric (or topological) space with countable basis and . Then, if is an open cover for then admits a countable subcover.
Proof: For each there exists a neighborhood containing it. But, since is a basis we may find some basic open set for which . Evidently then is an open cover for , and since it's a subset of it must be countable. Thus, for each element of choosing some for which produces a countable subcover for as desired.
So, we now claim that , as described in the problem, is perfect. To sync our notations call (the derived set) the set of all limit points of . Thus, we aim to show that .
So, to first see that let . Then, for each neighborhood of we have that there is some . But, since is a neighborhood of we see that is uncountable. Thus, since was arbitrary it follows that .
Conversely, suppose that then, there exists a neighborhood of for which every element of has a neighborhood for which clearly then is an open cover for . Thus, by our lemma it must admit a countable subcover . Note though that (this last part since by construction is countable and the countable union of countable sets is countable). and thus and thus . Thus, .
Thus, putting these two together gives .
So, now for the second part of the problem let be the countable basis for (the set of all open balls with rational radii at points of ) and let . We claim that . To see this let , then for some , but evidently then is a neighborhood of for which and so . Conversely, let , then there is some neighborhood of for which , and since is a basis we man find some for which and so evidently and so and thus . It follows that as desired. Note though that as desired (this past part gotten since by construction for each and since we have that is countable).